e constant of 35.0 N m–1 oscillates on a
and ω is 26.46 rads-1. Determine
0 cm.
s
25
(b) The speed of the object when y = 1.00 ×10–2 m a
v = ω A2 − y2 U
U
v = 26.46 (4.002 x10−2 )2 − (1.002 x10−2 )2
= . −
(b) The kinetic energy of the object
( ) ( )K = 1 mω 2 A2 − y2 = 1 k A2 − y2
22
( )( ) ( )K = 1 (35.0) 4.00×10−2 2 − 3.00×10−2 2
2
K = 1.23×10−2 J
and the potential energy of the object
U = 1 kx2
2
( )U = 1 (35.0) 3.00×10−2 2
2
U = 1.58×10−2 J
26
EXAMPLE 7.1.5 :
The length of a simple pendulum is 75.0 cm and it is rele
angle 8 to the vertical. Frequency of the oscillation is 0
Calculate the pendulum’s bob speed when it passes thro
lowest point of the swing. (Given g = 9.81 ms-2)
Solution: L = 0.75 m; θ = 8
At the lowest point, the velocity of the pendulum’s bob is m
vmax = Aω where A = L sin 8
( )vmax = L sin 8 (2πf )
( )vmax = 0.75sin 8 (2π (0.576))
vmax = 0.378 m s−1
eased at an L
0.576 Hz. 8
ough the
A
m
−A O A
maximum hence
27
7.2 : Graphs of simple harm
Learning Outcome:
a) Analyze the following graphs:
i. displacement-time;
ii. velocity-time;
iii. acceleration-time; and
iv. energy-displacement.
monic motion
:
28
7.2 (a) Analyse graphs of simple harmonic motion
i) Graph of displacement-time (y-t)
y = Asin (ωt )
y(m) Period
Amplitude
0 TT 3T T t(s)
42 4
−
ii) Graph of velocity-time (v-t)
v = Aω cos(ωt)
v(ms −1 )
0 T T 3T T t(s)
4 24
−
29
7.2 (a) Analyse graphs of simple harmonic motion
iii) Graph of acceleration-time (a-t)
a = − Aω 2 sin (ωt)
a(ms−2 )
0 T T 3T T t(s)
− 4 24
iv) Graph of energy-displacement (E-y)
E(J ) E = 1 mω 2 A2 = constant
2
U = 1 mω 2 y2
2
( )K = 1 mω2 A2 − y2
2
30
EXAMPLE 7.2.1 :
The displacement of an oscillating object as a function of tim
(cm) From
a. the
15.0 b. the
c. the
0
0.8 t(s) d. the
− 15.0 (b) The angular
Solution: ω=2
(a) From the graph: ω=
Amplitude, A = 0.15 m
Period, T = 0.8 s
Frequency,
f=1= 1
T 0.8
f = 1.25 Hz
me is shown below.
m the graph above, determine for these oscillations
e amplitude, the period and the frequency,
e angular frequency,
e equation of displacement as a function of time,
e equation of velocity and acceleration as a function of time.
r frequency of the oscillation is given by
2π = 2π
T 0.8
2.5π rad s−1
31
(c) The angular frequency of the oscillation is given by
By applying the general equation of displacement in SH
y = Asin (ωt )
= ( . ) where y and t are measured in
(d) i. The equation of velocity as a function of time is
v = Aω cos(ωt)
v = 0.15(2.5π )cos 2.5πt
v = 0.375π cos 2.5πt where is in m s−1 and
ii. and the equation of acceleration as a function of tim
= − 2 sin
= −0.15 2.5 2 sin 2 . 5
a = −0.938π 2 sin 2.5πt where is in m s−2 and
HM
n cm and t in second.
is in seconds.
me is
d is in seconds.
32
EXAMPLE 7.2.2 :
The displacement x, of a particle in simple harmonic motion v
= ( )
Where y is in m and t is in second. Sketch the graph of:
a) Displacement against time
b) Velocity against time
c) Acceleration against time
Solution:
a) = ( ) y(m)
A
0
= 2 −
= 2 = .
10
varies with time, t according to the equation
. . t (s)
33
b) = 2 sin(10 )
= cos
= (2) 10 cos (10 )
= 20 cos (10 )
( −1)
0 . . t(s)
−
c) = − 2 sin
= −(2) 10 2 sin (10 )
= −200 2 sin (10 )
( −2)
0 . . t(s)
−
34
EXAMPLE 7.2.3 :
A 2.0 kg body oscillate in simple harmonic motion. If its kinetic
energy, K changes with displacement, x as shown in figure, find
a) The amplitude
b) Period
c) The maximum acceleration
Solution:
a) The amplitude : A =0.2m
b) Period
( )K = 1 mω2 A2 − y2 c) maxim
2
Find first. Substitute y=0
2.0 = 1 (2) 2 (0.2)2−02
2
= 7.071 −1
= 2 = 2 = .
7.071
d
y (m)
−
mum acceleration,
= 2
= (0.2)(7.071)2
= (0.2)(7.071)2
= −
35
7.3 : Period of Simple Harmo
Learning Outcome:
a) Use expression for period o
mass-spring system.
(i) simple pendulum:
T = 2π l
g
(ii) single spring:
T = 2π m
k
onic Motion
of SHM, T for simple pendulum and
36
7.3 (i) Simple Pendulum
T =2
θL whe
m :
:
:
The conditions for a simple pendulum
i. the angle, θ has to be small (le
ii. the string has to be inelastic and
iii. only the gravitational force and t
acting on the simple pendulum.
2π L
g
ere,
period of the simple pendulum
length of the string
acceleration due to gravity
m to execute SHM :
ess than 10°).
d light.
tension in the string
37
7.3 (ii) Mass-spring system
=
F F1
x O a whe
O
m :
m :
mg :
mg
The conditions
i. The elastic
the spring
ii. The spring
iii. No air resis
= 2
ere,
period of the spring−mass system
: mass of the object
spring constant (force constant)
s for the spring-mass system executes SHM:
c limit of the spring is not exceeded when
g is being pulled.
is light and obeys Hooke’s law.
stance and surface friction.
38
EXAMPLE 7.3.1 :
What is the acceleration due to gravity in a region wh
75.0 cm has a period of 1.7357 s?
Solution:
T = 2π l
g
T 2 = 4π 2 l
g
( )g
= 4π 2l = 4π 2 75×10−2
T2 (1.7357)2
g = 9.83ms−2
here a simple pendulum having a length
39
EXAMPLE 7.3.2 :
When a family of four with total mass of 200 kg step into t
a) What is the spring constant of the car’s springs, assumin
b) Determine the period and frequency of the car after hit
the shock absorbers are poor, so the car really oscillates up
Solution:
a) Given , =1200 kg, 10−2
mass of family, =200 kg , y= 3 ×
m
� = 0
− = 0
Fs= -ky =
y =
= (
=
= . ×
their 1200-kg car, the car’s springs compress 3.0 cm.
ng they act as a single spring?
tting a bump. Assume
p and down.
b) = 2
= 2 (200 + 1200)
(6.54 × 104)
= .
= 1
(200)(9.81)
(3 × 10−2) 1
= (0.92)
× −
= . 40
7.4 : Properties of waves
Learning Outcome:
a) Define wavelength.
b) Define and use wave number, =
c) Solve problems related to equation
d) Discuss and use particle vibrationa
e) Use particle vibrational velocity,
f) Use wave propagation velocity, =
g) Analyse the graphs of:
i. displacement–time, y-t
ii. displacement–distance, y-x
2
λ
n of progressive waves, , = sin( ± )
al velocity and wave propagation velocity
= cos( − )
=
41
7.4 (a) Define wavelength
PROGRESSIVE WAVES :
a) Defined as the waves propagated continuously from
b) The progressive waves have a definite speed called
c) The direction of the wave speed is always in the sam
Figure below shows a periodic
sinusoidal waveform.
λ
BC
SPλ Q
λ T
m a source of disturbance
the speed of propagation or wave speed.
me direction of the wave propagation.
Sinusoidal Wave Parameters
1) Wavelength, λ 42
Defined as the distance between two
consecutive particles (points) which have the
same phase in a wave.
The S.I. unit of wavelength is meter (m).
Wavelength can be measured when:
i. Particle B is in phase with particle C.
ii. Particle P is in phase with particle Q
iii. Particle S is in phase with particle T
Sinusoidal Wave Parameters
(displacement)
Ay P
Ox
−A
2) Displacement, y
Defined as the distance moved by a particle f
along a wave.
3) Amplitude, A
Defined as the maximum displacement from
trough of the wave motion.
v
(distance from origin)
from its equilibrium position at every point
m the equilibrium position to the crest or
43
(displacement)
A
O
− A
4) Period, T
• Defined as the time taken for a particle (
this period, T the wave profile moves a d
• Its unit is second (s).
5) Frequency, f
T=1 • Defined as the number of cyc
f second.
• Its unit is hertz (Hz) or s−1.
v
( )
(point) in the wave to complete one cycle. In
distance of one wavelength, λ.
cles (wavelength) produced in one
44
7.4 (b) Define and use wave number,
6) Wave number, k
• Defined as the number of waves in a unit
• The S.I. unit of wave number is m−1.
= 2
tdistance.
45
EXAMPLE 7.4.1 :
A sinusoidal wave traveling in positive x direction ha
cm and a frequency of 8.0 Hz. Find the wave numbe
the wave.
Solution:
∶ = 2 = 2 = 15.71 −1
λ 0.4
∶ = 1 = 1 = 0.125
8
∶ = 2 = 2 = 50.27 −1
0.125
= λ = 8 0.4 = 3.2 −1
as an amplitude of 15.0 cm, a wavelength of 40.0
er k, period T, angular frequency ω & speed v of
46
EXAMPLE 7.4.2 :
A student reading his physics book on a lake dock notices
crests is 0.75 m, and he then measures the time of arrival b
(a) the frequency
(b) the speed of the waves ?
Solution: Given: = 0.75 m ; T = 1.6 s
(a) = 1 1 = 0.625s
= 1.6
(b) = 0.75 = 0.47m s−1
= 1.6
that the distance between two incoming wave
between the crests to be 1.6 s. What are
47
7.4 (c) Solve problems related to equation of p
The general wave equation for a sinusoidal p
(1) moves to the right ( in + x direction
y(x,t) = Asin (ω t − k x)
(2) moves to the left ( in – x direction )
y(x,t) = Asin (ω t + k x)
where
y : displacement of the particle from
A : amplitude
ω : angular frequency
k : wave number
t : time
progressive wave, ( , ) = sin( ± )
progressive wave that :
)
mequilibrium position
48
EXAMPLE 7.4.3 :
A progressive wave is described as = 2 sin 2 +
0.4 80
the following from this wave:
a) Amplitude b)wavelength
c) Frequency d) speed
Solution:
(a) =
(b) From equation given, we know that: = 2 where
80
λ= 2 =
2
80
(c) From equation given, we know that: = 2 where
0.4
2
0.4
= 2 = .
, where x and y are in cm and t is in seconds. Determine
e = 2 (d) = λ
λ = 2.5 (0.8)
= −
e = 2
49
7.4 (d) Distinguish between particle vibrationa
• A transverse wave on a str
constant speed, vwave.
• A string particle moves up
undisturbed position of th
• The speed of the particle
as the wave passes.